Question

A bar magnet of length $14cm$ is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of $18cm$ from the center of the magnet. If $B_H=0.4G$, the magnetic moment of the magnet is $\left(1G=10^{-4}T\right)$

• A. $2.880\times 10^{3}J{T}^{-1}$
• B. $2.880\times 10^{2}J{T}^{-1}$
• C. $2.880J{T}^{-1}$
• D. $28.80J{T}^{-1}$

Answer 1

Answer: (C)

i.e. $\frac{2{\mu }_0}{4\pi }\frac{m}{r^2}\times \frac{7}{r}=0.4\times 10^{-4}$
$\Rightarrow 2\times 10^{-7}\times \frac{m\times 7}{{\left(7^2+18^2\right)}^{3/2}}\times 10^4$
$=0.4\times 10^{-4}$
$m=\frac{4\times 10^{-2}\times (373{)}^{3/2}}{14}$
$M=m\times 14cm=m\times \frac{14}{100}$
$=\frac{0.04\times (373{)}^{3/2}}{14}\times \frac{14}{100}$
$=4\times 10^{-4}\times 7203.82=2.88J/T$